# If $T$ is a symmetric bilinear form on vector space $V$, and let $U$ be a finite dimensional subspace of $V$, then $V=U+U^{\bot}$

Here is the full question:

If $$T$$ is a symmetric bilinear form on vector space $$V$$, and let $$U$$ be a finite dimensional subspace of $$V$$, then $$V=U+U^{\bot}$$,

where $$U^{\bot}$$ is the orthogonal complement for $$U$$, given by $$U^{\bot}=\{v \in V \mid T(u,v)=0\}$$.

How does one prove that $$V=U+U^{\bot}$$?

We haven't introduced inner product spaces in class yet, and I understand that if $$T$$ was the inner product, standard proofs use: assume if $$V \neq U + U^{\bot}$$, there exists a vector $$\alpha$$ such it is orthogonal for all $$\beta \in \text{span}(U,U^{\bot})$$, implies $$<\alpha, \beta>=0$$, which implies $$\alpha \in U^{\bot}$$, contradiction.

But in this case, I cannot assume that $$T$$ is not an inner product? Are there any hints on how I can prove this statement?

• You need more conditions on your bilinear form. The question as currently stated is not true, as the zero bilinear form and $U = V$ gives counter example. – WhatsUp Mar 20 '20 at 3:53
• If T wasn't the zero bilinear forms, are there any counterexample? What further conditions do you need? Are those the axioms of inner product spaces? – Yip Jung Hon Mar 20 '20 at 4:41
• This is not true even if you assume that $T$ is non-degenerate (e.g. take some lightlike plane in Minkowski space). You do get at least that the dimensions of $U$ and $U^\perp$ add up to the dimension of $V$, provided either $T$ or $T|_U$ is non-degenerate. – Ivo Terek Mar 20 '20 at 7:11

## 1 Answer

I believe this is not true in general. Take $$A=\left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right)$$ and define $$T(u,v)=u^T A v$$, the symmetric bilinear form.

Let $$W=\text{span}\{(1,-1,0)\}$$. Then the only vectors that fulfils $$(a,b,c)\left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right) \left(\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right)$$

is such that $$(a,b,c) \left(\begin{array}{r} 0 \\ 0 \\ 1 \end{array}\right)=0$$ has dimension 2, but since every vector in $$W$$ satisfies this equation as well, $$W \subset W^{\bot}$$ and their sum cannot be possibly equal to $$V$$.